Let X and Y be sets and f: X - > Y a map. For any subset A subset X, recall that we define the
image of A by f (A) := {y E Y : y = f(a) for some a E A}. Assume A subset X arid B subset X
and show that f(A u B) = f(A) u f(B). Bonus: (5 points) Under the same conditions is f(A n B) =
f(A) n f(B)?
Solution
X and Y are two sets and f: x -->Y
A is a subset of X and B is a subset of x
Obviously AUB is a subset of X
AUB = A-B U A int B U B-A
Hence f(AUB) = f(A-B) Uf(Aint B)Uf(B_A)
= f(A) Uf(B-A) as A-B , A and B-A are disjoint sets
= f(A) U f(B)
f(A int B) = {b: b = f(a) where a is in A int B)
f(A) int f(B) = {b = f(a) such that a is in both A and B
Hence both are equal..
Beginners Guide to TikTok for Search - Rachel Pearson - We are Tilt __ Bright...
Let X and Y be sets and f X - Y a map. For any subset A subset X,.pdf
1. Let X and Y be sets and f: X - > Y a map. For any subset A subset X, recall that we define the
image of A by f (A) := {y E Y : y = f(a) for some a E A}. Assume A subset X arid B subset X
and show that f(A u B) = f(A) u f(B). Bonus: (5 points) Under the same conditions is f(A n B) =
f(A) n f(B)?
Solution
X and Y are two sets and f: x -->Y
A is a subset of X and B is a subset of x
Obviously AUB is a subset of X
AUB = A-B U A int B U B-A
Hence f(AUB) = f(A-B) Uf(Aint B)Uf(B_A)
= f(A) Uf(B-A) as A-B , A and B-A are disjoint sets
= f(A) U f(B)
f(A int B) = {b: b = f(a) where a is in A int B)
f(A) int f(B) = {b = f(a) such that a is in both A and B
Hence both are equal.
